Answer: A, B, C
In the first three, Br⋅ adds to the carbon that gives the more stable radical (terminal carbon for propene and 2-methylpropene, either carbon for symmetrical but-2-ene), so they are correct. For styrene, Br⋅ adds to the terminal carbon giving the benzylic radical, so the product is CX6HX5CHX2CHX2Br, not the Markovnikov isomer.